Question

How do you solve $ \left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)=24 $ ?

Answer 1

Hint:
We will rearrange the factors in the given expression and multiply them in pairs. After that, we will obtain two factors with some similar terms. We will substitute this similar part with a different variable. Then we will obtain a quadratic equation in the new variable. We will solve this equation to obtain two solutions for the new variable. Then using these two solutions, we will find the solutions for the original variable.

Complete step by step answer:
The given expression is the following,
 $ \left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\left( x-4 \right)=24 $
Now, let us rearrange the factor in the left hand side in the following manner,
 $ \left( x-1 \right)\left( x-4 \right)\left( x-2 \right)\left( x-3 \right)=24 $
We will multiply the first two factors and the next two factors separately. So, we get the following equation,
 $ \begin{align}
  & \left( {{x}^{2}}-x-4x+4 \right)\left( {{x}^{2}}-2x-3x+6 \right)=24 \\
 & \therefore \left( {{x}^{2}}-5x+4 \right)\left( {{x}^{2}}-5x+6 \right)=24 \\
\end{align} $
In the above equation, we have two factors and the first two terms in each factor is same. Now, we will substitute the similar part of both factors with a new variable. Let us assume that $ {{x}^{2}}-5x=y $ . So, the above equation becomes the following,
 $ \left( y+4 \right)\left( y+6 \right)=24 $
Now, let us simplify the above equation. We get the following equation,
 $ \begin{align}
  & {{y}^{2}}+4y+6y+24=24 \\
 & \therefore {{y}^{2}}+10y=0 \\
\end{align} $
We can factorize the above equation as
 $ y\left( y+10 \right)=0 $
Therefore, we have $ y=0 $ or $ y+10=0 $ . This means that $ {{x}^{2}}-5x=0 $ or $ {{x}^{2}}-5x+10=0 $ .
Let us consider the first case, where $ {{x}^{2}}-5x=0 $ . Solving this equation by the method of factorization, we get
 $ x\left( x-5 \right)=0 $
Therefore, we have $ x=0 $ or $ x-5=0 $ . This implies that $ x=0 $ and $ x=5 $ are two real solutions of the given equation.
Next, let us consider the second case, where $ {{x}^{2}}-5x+10=0 $ . We will use the quadratic formula to find the solution of this equation. The quadratic formula is given as
 $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
Substituting the appropriate coefficients in the above formula, we get
 $ \begin{align}
  & x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 10 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-40}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{-15}}{2} \\
 & \therefore x=\dfrac{5\pm \sqrt{15}i}{2} \\
\end{align} $
Therefore, $ x=\dfrac{5+\sqrt{15}i}{2} $ and $ x=\dfrac{5-\sqrt{15}i}{2} $ are two complex solutions of the given equation.
So, the solutions of the given equation are $ x=0 $ , $ x=5 $ , $ x=\dfrac{5+\sqrt{15}i}{2} $ and $ x=\dfrac{5-\sqrt{15}i}{2} $ .

Note:
It is important to know the methods to solve the quadratic equations. We can solve the quadratic equations using the quadratic formula, method of factorization or the method of completing the square. We should choose the method according to our convenience and ease of calculation. The important part in this question is to realize that rearranging the factors of the given equation makes it easier to solve the equation.